{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 3 : Fundamental concepts of Stoichiometry"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.1 page no : 41"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables \n",
"\n",
"m = 1000 * 0.4536; #kg/min pounds\n",
"M = 30.24; #gm/mol avg. molecular weight\n",
"\n",
"# Calculation \n",
"m1 = m * 60 / M;\n",
"\n",
"# Result\n",
"print \"molar folw rate = \",m1,\"kmol/hr\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"molar folw rate = 900.0 kmol/hr\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.2 page no : 42"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables \n",
"MK = 39.1; #potassium\n",
"MC = 12.0; # carbon\n",
"MO = 16.; # oxygen\n",
"\n",
"# Calculation \n",
"MK2CO3 = MK * 2 + MC + MO * 3;\n",
"m = 691.;\n",
"N = m / MK2CO3;\n",
"A = 6.023 * 10**23;\n",
"molecules = N * A;\n",
"\n",
"# Result \n",
"print \"Total no. of molecules = %.4e molecules\"%molecules\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Total no. of molecules = 3.0115e+24 molecules\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.3 page no : 43"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables \n",
"Na = 23.; #gm/mol weight of sodium\n",
"MNa = 100.; #kg sodium\n",
"\n",
"# Calculation \n",
"N = MNa * 1000 / Na ; #g-atoms \n",
"NNa2SO4 = N / 2;\n",
"\n",
"# Result \n",
"print \"(a) moles of sodium sulphate = %.3e mol\"%NNa2SO4\n",
"MNa2SO4 = 142.06;\n",
"m = NNa2SO4 * MNa2SO4/1000;\n",
"print \"(b)kilograms of sodium sulphate = %.2f kg\"%m\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a) moles of sodium sulphate = 2.174e+03 mol\n",
"(b)kilograms of sodium sulphate = 308.83 kg\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.4 page no : 43"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables \n",
"MFe = 55.85; \n",
"MO = 16.;\n",
"MS = 32.;\n",
"MFeS2 = MFe + MS * 2; # molecular weight of FeS2\n",
"MFe2O3 = MFe * 2 + MO * 3; # molecular weight of Fe2O3\n",
"MSO3 = MS + MO * 3; # molecular weight of SO3\n",
"m1SO3 = 100.; #kg\n",
"\n",
"# Calculation \n",
"N1 = m1SO3 / (MSO3); #kmol\n",
"NFeS2 = N1 / 2;\n",
"mFeS2 = NFeS2 * MFeS2;\n",
"\n",
"# Result\n",
"print \"mass of pyrites to obtain 100kg of SO3 = %.2f kg\"%mFeS2\n",
"m2SO3 = 50.; #kg\n",
"N2 = m2SO3 / (MSO3); #kmol\n",
"NO2 = N2 * 15/8.;\n",
"mO2 = NO2 * MO * 2;\n",
"print \"mass of Oxygen consumed to produce 50kg of SO3 = %.2f kg\"%mO2\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"mass of pyrites to obtain 100kg of SO3 = 74.91 kg\n",
"mass of Oxygen consumed to produce 50kg of SO3 = 37.50 kg\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.5 page no : 44"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables \n",
"MKClO3 = 122.55 # weight of potassium\n",
"mKClO3 = 100.; #kg potassium\n",
"\n",
"# Calculation \n",
"NKClO3 = mKClO3 / MKClO3;\n",
"NO2 = 3 * NKClO3 / 2;\n",
"V1 = 22.4143; #m**3/kmol;\n",
"V = V1 * NO2;\n",
"\n",
"# Result \n",
"print \"volume of oxygen produced = %.2f m**3\"%V\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"volume of oxygen produced = 27.43 m**3\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.6 page no : 44"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# Variables \n",
"mH2 = 100.; #kg hydrogen\n",
"\n",
"# Calculation \n",
"NH2 = mH2/2.016;\n",
"NFe = 3. * NH2 / 4;\n",
"mFe = NFe * 55.85;\n",
"\n",
"# Result \n",
"print \"(a)mass of iron required = %.2f kg\"%mFe\n",
"NH2O = NH2 \n",
"mH2O = NH2O * 18;\n",
"print \"mass of steam required = %.1f kg\"%mH2O\n",
"V1 = 22.4143; #m**3/kmol;\n",
"V = V1 * NH2;\n",
"print \"(b)Volume of hydrogen = %.1f m**3\"%V\n",
"\n",
"# Answer may vary because of rounding error.\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)mass of iron required = 2077.75 kg\n",
"mass of steam required = 892.9 kg\n",
"(b)Volume of hydrogen = 1111.8 m**3\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.7 page no : 46"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# variables \n",
"MCaCO3 = 100.08; # molecular weight of CaCO3\n",
"\n",
"# Calculation \n",
"GE = MCaCO3 / 2;\n",
"\n",
"# Result\n",
"print \"Gram equivalent wt. of CaCO3 =\",GE,\"g\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Gram equivalent wt. of CaCO3 = 50.04 g\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.8 page no : 48\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# variables \n",
"m1 = 1.; #kg (mass in air)\n",
"m2 = 0.9; #kg (mass in water)\n",
"m3 = 0.82; #kg (mass in liquid)\n",
"\n",
"# Calculation \n",
"L1 = m2 - m1; #kg (loss of mass in water)\n",
"L2 = m3 - m1; #kg (loss of mass in liquid)\n",
"sp_g = L2 /L1;\n",
"\n",
"# Result\n",
"print \"specific gravity of liquid = \",sp_g\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"specific gravity of liquid = 1.8\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.9 page no : 49\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# variables \n",
"m1 = 10. #kg liquid A\n",
"m2 = 5. #kg liquid B\n",
"sp_g1 = 1.17; # gravity A\n",
"sp_g2 = 0.83; # gravity B\n",
"Dwater = 1000. #kg/m**3 water\n",
"\n",
"# Calculation \n",
"DA = Dwater * sp_g1;\n",
"DB = Dwater * sp_g2;\n",
"V1 = m1 / DA;\n",
"V2 = m2 / DB;\n",
"V = V1 + V2;\n",
"Dmix = (m1 + m2)/ V ;\n",
"sp_g3 = Dmix \n",
"\n",
"# Result\n",
"print \"specific gravity of mixture = %.f kg/m**3\"%sp_g3\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"specific gravity of mixture = 1029 kg/m**3\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.10 page no : 49"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# variables \n",
"Tw = 100.; #Tw baume scale\n",
"\n",
"# Calculation \n",
"sp_g = Tw/200 + 1;\n",
"Be = 145 - 145/sp_g;\n",
"\n",
"# Result\n",
"print \"specific gravity on beume scale = %.1f Be\"%Be\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"specific gravity on beume scale = 48.3 Be\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.11 page no : 49\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# variables \n",
"API1 = 30. #API gas oil\n",
"sp_g1 = 141.5/(131.5 + API1) # (since, API = 141.5/sp_g -131.5) gravity scale\n",
"Dwater = 999. #kg/m**3; density of water\n",
"\n",
"# Calculation \n",
"Doil1 = sp_g1 * Dwater;\n",
"V1 = 250.; #m**3\n",
"m1 = V1 * Doil1;\n",
"API2 = 15.; #API\n",
"sp_g2 = 141.5/(131.5 + API2); # (since, API = 141.5/sp_g -131.5)\n",
"Dwater = 999.; #kg/m**3;\n",
"Doil2 = sp_g2 * Dwater;\n",
"V2 = 1000.; #m**3\n",
"m2 = V2 * Doil2;\n",
"Dmix = (m1 + m2)/(V1 + V2);\n",
"\n",
"# Result\n",
"print \"density of the mixture = %.f kg/m**3\"%Dmix\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"density of the mixture = 947 kg/m**3\n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.12 page no : 51"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# variables \n",
"m1 = 250. #kg wet ammonium sulphate\n",
"mwater1 = 50. #kg moisture\n",
"\n",
"# Calculation \n",
"mdrysolid1 = m1 - mwater1;\n",
"wfe1 = mwater1 / m1;\n",
"wr1 = mwater1 / mdrysolid1; \n",
"wtpercentw1 = mwater1 * 100 / m1;\n",
"wtpercentd1 = mwater1 * 100 / mdrysolid1;\n",
"a = 90. #%\n",
"mwater2 = mwater1 * (1 - a/100);\n",
"m2 = mdrysolid1 + mwater2;\n",
"wfe2 = mwater2 / m2;\n",
"wr2 = mwater2 / mdrysolid1; \n",
"wtpercentw2 = mwater2 * 100 / m2;\n",
"wtpercentd2 = mwater2 * 100 / mdrysolid1;\n",
"\n",
"# Result\n",
"print \"(a)weight fraction of water at entrance =\",wfe1\n",
"print \"weight fraction of water at exit = %.3f\"%wfe2\n",
"print \"(b)weight ratio of water at entrance = \",wr1\n",
"print \"weight ratio of water at exit = \",wr2\n",
"print \"weight percent of moisture on wet basis at entrance = \",wtpercentw1,\"%\"\n",
"print \"(c)weight percent of moisture on dry basis at entrance = \",wtpercentd1,\"%\"\n",
"print \"weight percent of moisture on wet basis at exit = %.2f %%\"%wtpercentw2\n",
"print \"(d)weight percent of moisture on dry basis at exit = \",wtpercentd2,\"%\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)weight fraction of water at entrance = 0.2\n",
"weight fraction of water at exit = 0.024\n",
"(b)weight ratio of water at entrance = 0.25\n",
"weight ratio of water at exit = 0.025\n",
"weight percent of moisture on wet basis at entrance = 20.0 %\n",
"(c)weight percent of moisture on dry basis at entrance = 25.0 %\n",
"weight percent of moisture on wet basis at exit = 2.44 %\n",
"(d)weight percent of moisture on dry basis at exit = 2.5 %\n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.13 page no : 53"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# variables \n",
"mdrysolid = 100. #kg dry solids\n",
"percentin = 25. # water in the feed\n",
"\n",
"# Calculation \n",
"mwaterin = mdrysolid * percentin / 100;\n",
"percentout = 2.5;\n",
"mwaterout = mdrysolid * percentout / 100;\n",
"mremoved = mwaterin - mwaterout;\n",
"percentremoved = mremoved *100 / mwaterin ;\n",
"\n",
"# Result\n",
"print \"percentage of water removed = \",percentremoved\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"percentage of water removed = 90.0\n"
]
}
],
"prompt_number": 21
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.14 page no : 53"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# variables \n",
"m = 1. #kg wet ammonium sulphate\n",
"percent1 = 20. #% water\n",
"\n",
"# Calculation \n",
"mwaterin = m * percent1 / 100.\n",
"mdrysolid = m - mwaterin;\n",
"percent2 = 2.44; #%\n",
"mout = mdrysolid / (1 - percent2/100);\n",
"mwaterout = mout - mdrysolid;\n",
"mremoved = mwaterin - mwaterout;\n",
"percentremoved = mremoved * 100 / mwaterin ;\n",
"\n",
"# Result\n",
"print \"weight of water removed = %.2f kg\"%mremoved\n",
"print \"percentage of water removed = %.f %%\"%percentremoved\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"weight of water removed = 0.18 kg\n",
"percentage of water removed = 90 %\n"
]
}
],
"prompt_number": 23
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.15 page no : 55"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# variables \n",
"mwater = 100. #kg water\n",
"mNaCl = 35.8 #kg solubility\n",
"\n",
"# Calculation \n",
"msolu = mwater + mNaCl;\n",
"mfr = mNaCl / msolu;\n",
"mpr = mfr * 100;\n",
"MNaCl = 58.45; #kg/kmol\n",
"NNaCl = mNaCl / MNaCl;\n",
"MH2O = 18. #kg/kmol\n",
"NH2O = mwater / MH2O;\n",
"Mfr = NNaCl / (NNaCl + NH2O);\n",
"Mpr = Mfr * 100;\n",
"N = NNaCl *1000 / mwater;\n",
"\n",
"# Result\n",
"print \"(a)mass fraction of NaCl = %.4f\"%mfr\n",
"print \"mass percent of NaCl= %.2f %%\"%mpr\n",
"print \"(b)mole fraction of NaCl = %.4f\"%Mfr\n",
"print \"mole percent of NaCl = %.2f %%\"%Mpr\n",
"print \"kmol NaCl per 1000 kg of water = %.3f kMol NaCl\"%N\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)mass fraction of NaCl = 0.2636\n",
"mass percent of NaCl= 26.36 %\n",
"(b)mole fraction of NaCl = 0.0993\n",
"mole percent of NaCl = 9.93 %\n",
"kmol NaCl per 1000 kg of water = 6.125 kMol NaCl\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.16 page no : 55"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# variables \n",
"Y = 0.015; #kg water vapour/kg dry air\n",
"Mair = 29.; #kg/kmol weight of air\n",
"Mwater = 18.016; #kg/kmol \n",
"\n",
"# Calculation \n",
"Nwater = Y / Mwater; #kmol\n",
"Nair = 1 / Mair; #kmol\n",
"Mpr = Nwater *100 / (Nwater + Nair);\n",
"Mr = Nwater / Nair;\n",
"\n",
"# Result\n",
"print \"(a)mole percent of water vapour = %.2f %%\"%Mpr\n",
"print \"(b) molal absolute humidity = %.4f kmol water/kmol dry air\"%Mr\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)mole percent of water vapour = 2.36 %\n",
"(b) molal absolute humidity = 0.0241 kmol water/kmol dry air\n"
]
}
],
"prompt_number": 26
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.17 page no : 56"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# variables \n",
"msolu = 100.; #g basis\n",
"MK2CO3 = 138.20; #g/mol molecular weight\n",
"percent1 = 50.; #% mass of K2CO3\n",
"\n",
"# Calculation \n",
"mK2CO3 = percent1 *msolu / 100;\n",
"NK2CO3 = mK2CO3 / MK2CO3;\n",
"mwater = msolu - mK2CO3;\n",
"Nwater = mwater / 18.06;\n",
"Mpr = NK2CO3 * 100 / (NK2CO3 + Nwater);\n",
"sp_gr =1.53;\n",
"Vsolu = msolu/sp_gr; #mL\n",
"Vwater = mwater / 1; #mL\n",
"Vpr = Vwater * 100/ Vsolu;\n",
"Molality = NK2CO3 / (mwater * 10**-3);\n",
"Molarity = NK2CO3 / (Vsolu * 10**-3);\n",
"Eq_wt = MK2CO3 / 2;\n",
"No = mK2CO3/Eq_wt;\n",
"N = No / (Vsolu * 10**-3);\n",
"\n",
"# Result\n",
"print \"(a)Mole prcent of salt = %.2f %%\"%Mpr\n",
"print \"(b)Volume percent of water = %.2f %%\"%Vpr\n",
"print \"(c)Molality = %.3f mol/kg\"%Molality\n",
"print \"(d)Molarity = %.3f mol/L\"%Molarity\n",
"print \"(e)Normality = %.2f N\"%N\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)Mole prcent of salt = 11.56 %\n",
"(b)Volume percent of water = 76.50 %\n",
"(c)Molality = 7.236 mol/kg\n",
"(d)Molarity = 5.535 mol/L\n",
"(e)Normality = 11.07 N\n"
]
}
],
"prompt_number": 27
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.18 page no : 58"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"# variables \n",
"msolu = 100.; #kg volume\n",
"percent1 = 60.; #% alcohol\n",
"Dwater = 998.; #kg/m**3 water\n",
"Dalco = 798.; #kg/m**3 alcohol\n",
"Dsolu = 895.; #kg/m**3 solution\n",
"\n",
"# Calculation \n",
"Vsolu = msolu/Dsolu;\n",
"malco = msolu * percent1 / 100;\n",
"Valco = malco / Dalco;\n",
"Vpr = Valco * 100 / Vsolu;\n",
"Malco = 46.048; #kg/kmol\n",
"N = malco/Malco;\n",
"Molarity = N/(Vsolu );\n",
"mwater = msolu - malco;\n",
"Molality = N * 1000 /mwater;\n",
"\n",
"# Result\n",
"print \"(a)Volume percent of ethanol in solution = %.1f %%\"%Vpr\n",
"print \"(b)Molarity = %.2f mol/L\"%Molarity\n",
"print \"(c)Molality = %.3f mol/(kg of water)\"% Molality\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)Volume percent of ethanol in solution = 67.3 %\n",
"(b)Molarity = 11.66 mol/L\n",
"(c)Molality = 32.575 mol/(kg of water)\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.19 page no : 63"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#CO + CL2 = COCl2\n",
"import math \n",
"# Variables \n",
"Np = 12.; #moles phosgene\n",
"NCl2 = 3.; #moles chlorine\n",
"NCO = 8.; #moles carbon monoxide\n",
"\n",
"# Calculation \n",
"N1Cl2 = NCl2 + Np;\n",
"N1CO = NCO + Np;\n",
"pr_ex = (N1CO - N1Cl2)* 100/N1Cl2;\n",
"pr_co = (N1Cl2-NCl2) * 100/ N1Cl2;\n",
"T = Np + NCl2 + NCO;\n",
"T1 = N1Cl2 + N1CO;\n",
"N = T / T1;\n",
"\n",
"# Result\n",
"print \"(a)percent excess of CO = %.2f %%\"%pr_ex\n",
"print \"(b)percent conversion = %.2f %%\"%pr_co\n",
"print \"(c)Moles of total products per mole of total reactants = %.3f\"%N\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)percent excess of CO = 33.33 %\n",
"(b)percent conversion = 80.00 %\n",
"(c)Moles of total products per mole of total reactants = 0.657\n"
]
}
],
"prompt_number": 29
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.21 page no : 64"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables \n",
"\n",
"msolu = 100.; #g mold of feed mixture\n",
"MK2CO3 = 138.20; #g/mol \n",
"ethylene = 53.89 # %\n",
"ethanol = 14.37 # %\n",
"ether = 1.8 # %\n",
"water = 23.35 # %\n",
"\n",
"# Calculation \n",
"Ethylene = ethylene * 83.57 / 100\n",
"Ethanol = ethanol*83.57 / 100\n",
"Ether = ether*83.57 / 100\n",
"Water = water*83.57 / 100\n",
"Inerts = 3 # mol\n",
"\n",
"conversion_ethylene = (60 - Ethylene)/60 * 100\n",
"yield_ethanol = Ethanol/(60 - Ethylene)*100\n",
"yeild_ether = Ether*2/(60-Ethylene) * 100\n",
"\n",
"# Result \n",
"print \"Conversation of ethylene = %.2f %%\"%conversion_ethylene\n",
"print \"Yield of ethanol = %.2f %%\"%yield_ethanol\n",
"print \"Yield of Ether = %d %%\"%yeild_ether\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Conversation of ethylene = 24.94 %\n",
"Yield of ethanol = 80.25 %\n",
"Yield of Ether = 20 %\n"
]
}
],
"prompt_number": 1
}
],
"metadata": {}
}
]
}